如何生成长度为8的倍数(对应于标准数据类型)的位集,其中每个位的概率为0或1?
最佳答案
以下作品。
以下代码完成了此任务:
#include <cstdint>
#include <iostream>
#include <random>
#include <algorithm>
#include <functional>
#include <bitset>
//Generate the goodness
template<class T>
T uniform_bits(std::mt19937& g){
std::uniform_int_distribution<T> dist(std::numeric_limits<T>::lowest(),std::numeric_limits<T>::max());
return dist( g );
}
int main(){
//std::default_random_engine can be anything, including an engine with short
//periods and bad statistical properties. Rather than cross my finers and pray
//that it'll somehow be okay, I'm going to rely on an engine whose strengths
//and weaknesses I know.
std::mt19937 engine;
//You'll see a lot of people write `engine.seed(std::random_device{}())`. This
//is bad. The Mersenne Twister has an internal state of 624 bytes. A single
//call to std::random_device() will give us 4 bytes: woefully inadequate. The
//following method should be slightly better, though, sadly,
//std::random_device may still return deterministic, poorly distributed
//numbers.
std::uint_fast32_t seed_data[std::mt19937::state_size];
std::random_device r;
std::generate_n(seed_data, std::mt19937::state_size, std::ref(r));
std::seed_seq q(std::begin(seed_data), std::end(seed_data));
engine.seed(q);
//Use bitset to print the numbers for analysis
for(int i=0;i<50000;i++)
std::cout<<std::bitset<64>(uniform_bits<uint64_t>(engine))<<std::endl;
return 0;
}
我们可以通过编译(
g++ -O3 test.cpp
)并通过以下方式进行一些统计来测试输出:./a.out | sed -E 's/(.)/ \1/g' | sed 's/^ //' | numsum -c | tr " " "\n" | awk '{print $1/25000}' | tr "\n" " "
结果是:
1.00368 1.00788 1.00416 1.0036 0.99224 1.00632 1.00532 0.99336 0.99768 0.99952 0.99424 1.00276 1.00272 0.99636 0.99728 0.99524 0.99464 0.99424 0.99644 1.0076 0.99548 0.99732 1.00348 1.00268 1.00656 0.99748 0.99404 0.99888 0.99832 0.99204 0.99832 1.00196 1.005 0.99796 1.00612 1.00112 0.997 0.99988 0.99396 0.9946 1.00032 0.99824 1.00196 1.00612 0.99372 1.00064 0.99848 1.00008 0.99848 0.9914 1.00008 1.00416 0.99716 1.00868 0.993 1.00468 0.99908 1.003 1.00384 1.00296 1.0034 0.99264 1 1.00036
由于所有的值(value)观都“接近”一个,我们得出结论,我们的使命已经完成。